There are two concentric metal shells of radii `r_(1)` and `r_(2) ( gt r_(1))`. If initially, the outer shell has a charge q and the inner shell is having zero charge and then inner shell is grounded. Find :
(i) Charge on the inner surface of outer shell.
(ii) Final charges on each sphere.
(iii) Charge flown through wire in the ground.

Answer

Let's solve the problem step by step.

### Given:
- Two concentric metal shells with radii \( r_1 \) (inner shell) and \( r_2 \) (outer shell, where \( r_2 > r_1 \)).
- Initial charge on the outer shell = \( q \).
- Initial charge on the inner shell = 0.
- The inner shell is grounded.

### (i) Charge on the inner surface of the outer shell

1. **Understanding the grounding of the inner shell**:
When the inner shell is grounded, it can exchange charge with the earth until its potential becomes zero.

2. **Let the charge that flows to the inner shell be \( Q \)**. Since the inner shell is grounded, its potential \( V_A \) must be zero. The potential at the inner shell due to its own charge and the charge on the outer shell can be expressed as:
\[ V_A = \frac{k \cdot Q}{r_1} + \frac{k \cdot q}{r_2} = 0 \]
where \( k \) is Coulomb's constant.

3. **Rearranging the equation**:
\[ \frac{k \cdot Q}{r_1} = -\frac{k \cdot q}{r_2} \]
\[ Q = -\frac{r_1}{r_2} \cdot q \]

4. **Charge on the inner surface of the outer shell**:
The inner surface of the outer shell will induce a charge equal in magnitude but opposite in sign to the charge on the inner shell:
\[ \text{Charge on inner surface of outer shell} = -Q = \frac{r_1}{r_2} \cdot q \]

### (ii) Final charges on each sphere

1. **Final charge on the inner shell**:
From the previous calculation, the final charge on the inner shell is:
\[ \text{Charge on inner shell} = Q = -\frac{r_1}{r_2} \cdot q \]

2. **Final charge on the outer shell**:
The outer shell initially had charge \( q \). The charge on the inner surface is \( -Q \), and the outer surface will have:
\[ \text{Charge on outer shell outer surface} = q + Q = q - \frac{r_1}{r_2} \cdot q = q \left(1 - \frac{r_1}{r_2}\right) \]

### (iii) Charge flown through the wire in the ground

1. **Initial charge on the inner shell** was 0, and after grounding, it became \( Q \):
\[ \text{Charge flown through the wire} = Q = -\frac{r_1}{r_2} \cdot q \]

### Summary of Results:
- (i) Charge on the inner surface of the outer shell: \( \frac{r_1}{r_2} \cdot q \)
- (ii) Final charge on the inner shell: \( -\frac{r_1}{r_2} \cdot q \); Final charge on the outer shell outer surface: \( q \left(1 - \frac{r_1}{r_2}\right) \)
- (iii) Charge flown through the wire in the ground: \( -\frac{r_1}{r_2} \cdot q \)